Question

Suppose that two hyperbolas with eccentricities \(e\) and \(E\) have perpendicular major axes and share a set of asymptotes. Show that \(e^{-2}+E^{-2}=1\)

Short answer

To prove the relationship between the eccentricities of two hyperbolas with perpendicular major axes that share a set of asymptotes, we first wrote the equations of the hyperbolas and their asymptotes. Then, we found the intersection of asymptotes and expressed the relationship in terms of eccentricities. The resulting expression is \(e^{-2}+E^{-2}=1\), where \(e\) and \(E\) are the eccentricities of the hyperbolas.

Step-by-step solution

Equation of the hyperbolas

The general equation of a hyperbola centered at the origin with major axis along the x-axis is \[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\] where \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. The eccentricity of the hyperbola is given by \[e=\frac{\sqrt{a^2+b^2}}{a}.\] Now, we have two hyperbolas, so let's denote their parameters by \((a_1,b_1,e_1)\) and \((a_2,b_2,e_2)\), with the first hyperbola having its major axis along the x-axis, and the second along the y-axis.

Equations of the asymptotes

The asymptotes of the hyperbolas are the lines that follow the equation \[\frac{x^2}{a^2}-\frac{y^2}{b^2}=0.\] For the two given hyperbolas, their equations will be \[\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=0\] and \[\frac{y^2}{a_2^2}-\frac{x^2}{b_2^2}=0.\]

Find the intersection of asymptotes

Since the hyperbolas share a set of asymptotes, their asymptote equations must be equal. Cross multiply and simplify: \[x^2b_2^2=a_1^2y^2 \quad \text{and} \quad x^2a_2^2=b_1^2y^2.\] Now let's take the ratio of these two equations: \[\frac{x^2b_2^2}{a_1^2y^2} = \frac{x^2a_2^2}{b_1^2y^2}.\] After cancelling x^2 and y^2, we get the following relationship: \[\frac{b_2^2}{a_1^2}=\frac{a_2^2}{b_1^2}.\]

Express relationship in terms of eccentricities

Firstly, let's denote our eccentricities E and e just as in the exercise. Let \(E = e_1\) and \(e = e_2\). We recall the relationship between the eccentricity and the semi-axes of a hyperbola: \[E=\frac{\sqrt{a_1^2+b_1^2}}{a_1}, \quad e=\frac{\sqrt{a_2^2+b_2^2}}{a_2}.\] Squaring these relationships, we obtain: \[E^2=\frac{a_1^2+b_1^2}{a_1^2}, \quad e^2=\frac{a_2^2+b_2^2}{a_2^2}.\] Now, we can rewrite these equations as \[b_1^2=a_1^2(E^2-1), \quad b_2^2=a_2^2(e^2-1).\] Substituting these into the relationship found in step 3: \[\frac{a_2^2(e^2-1)}{a_1^2}=\frac{a_1^2(E^2-1)}{a_2^2}.\] Now, we simply cancel out \(a_1^2a_2^2\) and rearrange to form the expression we are asked to prove: \[e^2E^2-1=(E^2-1)(e^2-1).\] \[1 - e^2E^2 = (1 - E^2)(1 - e^2).\] \[1=e^2-e^2E^2+E^2-E^2e^2.\] \[1=E^{-2}-1+e^{-2}.\] Finally, we get the desired expression: \[e^{-2}+E^{-2}=1.\]